In the figure above, the work done by the force of 100N inclined at an angle of 60o to the object dragged horizontally to a distance of 8m is
800J
600J
100J
400J
Correct answer is D
Since the force is inclined at an angle 60o to the horizontal, the effective component of te force along the horizontal is given by F x cos 60o
= 100 x 0.5 = 50N.
∴ work done = F x cos 60 x displacement
= 50 x 8.0m
= 400.0J
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